TL;DR - I suspect your confusion lies in the Physics 101 example that e.g. the ordered pair ("temperature","pressure") does not define a vector because when we change our coordinates, temperature and pressure don't transform. However, if we are working in cartesian coordinates, the object (temperature)$\hat x$ + (pressure)$\hat y$ is a linear combination of our basis vectors, and therefore does transform appropriately. This is just as well-defined as the vector $\mathbf V = 3\hat x + 4\hat y$.
In other words, if you write down a physically meaningless (but perfectly well-defined) vector field in some coordinate system, then it will change in the usual way when you move to a different coordinate system.
Here's the long answer:
So suppose we have a function $F$ from $\mathbb R^2$ to $\mathbb R^2$ defined by $F(x,y)=(g(x,y),h(x,y))$ where $g$ and $h$ represent temperature and pressure respectively (the point is, they are both scalar fields).
Okay, subtlety number one: Is the domain of $F$ the manifold $\mathcal M =\mathbb R^2$, or the image of $\mathcal M$ under a cartesian coordinate chart?
$$x :\mathcal M \rightarrow \mathbb R^2$$
$$(a,b) \mapsto (a,b)$$
Mathematical Interlude
Despite being one of the simplest possible manifolds, $\mathbb R^2$ is actually terrible from a pedagogical point of view precisely because it's so easy to get confused on this issue. The manifold $\mathcal M = \mathbb R^2$ is abstract; points $p\in \mathbb R^2$ consist of ordered pairs of real numbers $(a,b)$, but those numbers are not coordinates for $p$. We can introduce coordinates by defining a coordinate chart on some open neighborhood of $p$. For example, we might coordinatize the upper half-plane via the polar coordinate chart:
$$\pi : \mathbb R_+^2 \rightarrow \mathbb R \times(0,\pi)$$
$$(a,b) \mapsto \left(\sqrt{a^2+b^2},\sin^{-1}\left(\frac{b}{\sqrt{a^2+b^2}}\right)\right)$$
where the first coordinate is interpreted as the radial coordinate and the second as the angular coordinate.
Any function which is defined at the manifold level - e.g. some $f:\mathcal M \rightarrow \mathbb R$ - has a corresponding expression in each coordinate chart. For example, let $f:\mathcal M \rightarrow \mathbb R$ be defined by $(a,b)\mapsto a$. If we descend into the polar coordinate chart, we could consider the function
$$f_\pi: \mathbb R\times (0,\pi)$$
$$(r,\theta) \mapsto (f\circ \pi^{-1})(r,\theta) = f\big(r\cos(\theta),r\sin(\theta)\big) = r\cos(\theta)$$
$f_\pi$ is the expression of the (manifold-level) function $f$ in the $\pi$ coordinate chart. Changing to a different chart entails mapping points back to $\mathcal M$ via $\pi^{-1}$, then applying the new coordinate chart. For example, if we wanted to use the cartesian chart defined above, we would have
$$f_x = f\circ x^{-1} = f\circ \pi^{-1} \circ \pi \circ x^{-1} = f_\pi \circ (\pi\circ x^{-1})$$
The map $\pi \circ x^{-1}$ is called the chart transition map between the cartesian chart $x$ and the polar chart $\pi$; it is easily seen to be
$$\pi \circ x^{-1}: \mathbb R_+^2 \rightarrow \mathbb R\times(0,\pi)$$
$$(a,b) \mapsto \left(\sqrt{a^2+b^2},\sin^{-1}\left(\frac{b}{\sqrt{a^2+b^2}}\right)\right)$$
and so
$$f_x : (r,\theta)\in \mathbb R\times(0,\pi) \mapsto r\cos(\theta)$$
as anticipated.
End Interlude
The point of that somewhat length example is that when you say $F:\mathbb R^2\rightarrow \mathbb R^2$, its not clear whether you are defining an express at the manifold level - in which case there are no coordinates being used at all, and no transformations to consider - or at the level of (presumably cartesian) coordinates, in which case your $F$ is really $F_x \equiv F \circ x^{-1}$, and a change of chart is effected by simply inserting a chart transition map, e.g. $F_\pi \equiv F_x \circ (x\circ \pi^{-1})$.
From the viewpoint of differential geometry the above function can be seen as a (coordinate representation of a) vector field (i.e. a map from the manifold to the tangent bundle)
Okay. Based on this, I will assume we are working in cartesian coordinates. You are defining a vector field $\mathbf V$ on $\mathbb R^2$ whose $x$-component is the temperature and whose $y$-component is the pressure. That defines a little directional derivative which sits at each point, with the vector field being
$$\mathbf V = g(x,y)\frac{\partial}{\partial x} + h(x,y) \frac{\partial}{\partial y}$$
Mathematically, since this is a section of the projection map it is forced to obey the vector transformation law, yet physically it's intuitively clear that this is not a vector field but rather just a bunch of scalars so do we implement a vector transformation law or not (under a change of coordinates); does it even make sense to talk about?
I don't know what you mean here. It is a perfectly well-defined vector field. It doesn't have any physical significance, as far as I can tell, but that doesn't mean it isn't a vector field.
A change of coordinates induces a change of basis, so you're asking whether the components of $\mathbf V$ change when we go from the cartesian basis $\left\{\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right\}$ to e.g. the polar basis $\left\{\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}\right\}$, and the answer is obviously yes - the polar coordinate unit vectors generally point in different directions than the cartesian unit vectors, after all. If you replace
$$\frac{\partial}{\partial x}\mapsto \frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta}$$
$$\frac{\partial}{\partial y}\mapsto \frac{\partial r}{\partial y} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial y} \frac{\partial}{\partial \theta}$$
in our original expression for $\mathbf V$, then when the dust settles you'll have something of the form
$$\mathbf V = V^r \frac{\partial}{\partial r} + V^\theta \frac{\partial}{\partial \theta}$$
where $V^{r}$ and $V^\theta$ are some (position-dependent) linear combinations of $g$ and $h$. That's all the vector transformation rule is - expressing the same vector field using different basis vectors requires different components, which should be fairly obvious.
Actually, this isn't quite right - your functions $g$ and $h$ are really $g_x=g\circ x^{-1}$ and $h_x = h\circ x^{-1}$, so under change of chart they would also be replaced by $g_\pi = g_x \circ (x\circ \pi^{-1})$ and $h_\pi = h_x \circ (x\circ \pi^{-1})$.
More generally if we have an $n$-dimensional smooth manifold $\mathcal M$ (so think of a Riemannian 4-manifold for instance) and a function $F$ from $M$ to $\mathbb R^n$, will the physical nature of this function (i.e. depending on what physical quantity it represents) govern its transformation behavior?
No. As long as you have a well-defined function at the manifold level, that immediately translates into an expression in whatever coordinate chart you wish to work in. There is nothing transformative about this idea - if you have a point $p$ which is being mapped to some other space and you label $p$ by some coordinates, then you get a function which eats those coordinates. If you change coordinates, you change the function.
In this case, that other space was the tangent bundle, and we performed a corresponding chart transformation on that (i.e. a change of basis) which was induced by the chart transformation on the manifold $\mathcal M=\mathbb R^2$, which added a layer of complexity.
